Solution
We are given that
![\begin{gathered} p(A)=0.2 \\ p(B)=0.45 \\ p(C)=0.35 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/t5hrusb6h13ey3oc9gkbto4bowqkt4cjj7.png)
Note 1: Probability Formula To use
![p(A\cup B)=p(A)+p(B)-p(A\cap B)](https://img.qammunity.org/2023/formulas/mathematics/college/tck0phg2qjkk7w0qkpodhrs9w4wfh8deca.png)
Note 2: Team A, B and C are Mutually Exclusive
![\begin{gathered} p(A)+p(B)+p(C)=0.2+0.45+0.35=1 \\ Th\text{ey are mutually exclusive} \\ A\cap B=B\cap C=A\cap C=\varnothing \\ p(A\cap B)=p(B\cap C)=p(A\cap C)=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/taa3qt6xt7pfljtrb2juton8656eyoq30k.png)
Therefore, the formula to use now is
![p(A\cup B)=p(A)+p(B)](https://img.qammunity.org/2023/formulas/mathematics/college/zoljjgvl6g3qvbcsy5y4w2lzt7raze11w7.png)
For Anna
Anna can join either team A or team B.
We calculate the probability
![\begin{gathered} p(A\cup B)=p(A)+p(B) \\ p(A\cup B)=0.2+0.45 \\ p(A\cup B)=0.65 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/naty2811cnnl3wj9re9idjd473bezrxi18.png)
For Elina
Elina can join either team B or team C.
We calculate the probability
![\begin{gathered} p(B\cup C)=p(B)+p(C) \\ p(B\cup C)=0.45+0.35 \\ p(B\cup C)=0.8 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/t1q68q1z93ft3x0b2fpzcch70hn75o2014.png)
For Nancy
Nancy can join either team A or team C.
We calculate the probability
![\begin{gathered} p(A\cup C)=p(A)+p(C) \\ p(A\cup C)=0.2+0.35 \\ p(A\cup C)=0.55 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/l97vjy1mx90zuogd20notcme4rkrlhofcz.png)
The one with the highest probability is most likely to win and that is
ELINA
Correct answer is Elina